Drupal 8.x

I'd like to create a View that display content of two different types but one of which filtered by a taxonomy term, for example Bees.

Let say that I have the content type Articles associated to a vocabolary and the content type Basic Pages, I'd like to make a view with a list of all the basic pages and the Articles that have the taxonomy term Bees.

Is it possible to do that using the filter criteria?

after download drupal files on host, go to my site for continue install, but "Additional uncaught exception thrown while handling exception"

what does it mean? what will i do?

php 7.0, drupal 8.2.6

Additional uncaught exception thrown while handling exception.

Original

Error: Call to undefined function Drupal\Core\Site\hash_hmac() in Drupal\Core\Site\Settings::getApcuPrefix() (line 168 of /home/optstone/optstone.ru/docs/core/lib/Drupal/Core/Site/Settings.php).

I have pages Type "A" and pages Type "B".
Type B contain a field reference to content Type "A". (new field to categorize Type B parent)
I want to add a bloc into Type A, showing every page of Type B link to current page (children).

so, I create a new view/bloc, for the filter I choose "Content : Parent (field_parent_as_type_A)", but I dont know what I can use for the value. It work if I write the node number of one Type A page, but I want this bloc to be dynamic and use the current nid.

I'm trying to place a form in a custom block to be placed in a sidebar, but the form never shows, only a link.

Hi,

In some examples I saw, for validating one entity we must use two classes, for example in commerce product, it has ProductVariationSkuConstraint.php and ProductVariationSkuConstraintValidator.php for validating the SKU field. If I have several fields in one entity, is it okay to create only two classes for validating all the fields there? Or should I create those two classes for each fields, which means there will be a lot of classes for validation?